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anonymous
 one year ago
Suppose ΔABC is an acute triangle. Let the bases of the altitudes at B and C be E and F, respectively. Prove ΔABC and ΔAEF are similar
anonymous
 one year ago
Suppose ΔABC is an acute triangle. Let the bases of the altitudes at B and C be E and F, respectively. Prove ΔABC and ΔAEF are similar

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I typed the question verbatim, but it sounds odd that E and F would be bases. I would think they would be points. Regardless of that, still am unsure of how to do the problem.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437110964213:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, okay, so they really are supposed to be points E and F. Maybe I drew my picture wrong but, at least the way yours is drawn, you could maybe show FE is parallel to BC?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2That is equivalent, but there is a simpler way here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, thats equivalent to saying theyre similar? Saying those lines are parallel?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2First establish trivially that \(\triangle OFB\sim\triangle OEC\) by \(AA\) dw:1437112233723:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, made it that far at least xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Next show that \(\triangle AFC\sim AEB\) by AA dw:1437112450857:dw actually we don't need the first step :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2setup a proportion and we're done!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, I have that similarity. After that, maybe because I was being confused by my picture, I wasnt sure. And we're setting up a proportion based on the sides of triangle AFC and triangle AEB?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes : \[\dfrac{AF}{AE}=\dfrac{AC}{AB}\] together with \(\angle A\cong \angle A\) by reflexive property yield \(\triangle ABC\sim\triangle AEF\) by \(SAS\) similarity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, we didn't really talk about a lot of the high school geometry postulates and such. It seems like the professor just assumes that what we are given implies everything else. Maybe what we were given implies SAS similarity? This is all we were given about similar triangles: "ΔABC ~ ΔA'B'C' iff \(\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC}\)" Or "ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2"ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'" This is equivalent to AA similarity because the third angle is determined by triangle sum property once the first two angles are known.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2"ΔABC ~ ΔA'B'C' iff \(\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC}\)" This is SSS similarity, lets see if we can jump to this from SAS

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2here it is, pls scroll directly to SAS similarity http://ceemrr.com/Geometry1/SSSandSASsimilarity/SSSandSASsimilarity_print.html they're proving SAS similarity using AA

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alrighty, Ill take a look at that and make sure I see the SAS. Thanks again :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2This is analogous to SAS congruence : two triangles are congruent if the corresponding pairs of two adjacent sides and the included angle are congruent

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2SAS similarity says : two triangles are "similar" if the corresponding pairs of two adjacent sides are "proportional" and the included angles are congruent.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In the link, the statement they give for SAS says "If \(\frac{AB}{DE} = \frac{AC}{DF}\) and \(∠A\) congruent to \(∠B\)..." Yet in their diagram, it looks as if it would need to say that \(∠A\) congruent to \(∠D\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ahh, I overlooked that. thats definitely a typo yeah

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437114236422:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, cool. I gotta learn to use the latex and drawing tools and such like you know how to, haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I'm using a small chrome extension that gives plotting/graphing like features to the drawing tool
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